Question

What is the slope of `f(x)=xe^(x-x^2)` at `x=-1`?

Answer 1

The slope of the function at x=-1 is `-2/e^2`

Explanation:

In order to find the slope of a function,we have to firstly determine its derivative.
so,
`d/dx(xe^(x-x^2))`

Applying product rule,

`(f.g)'=f'.g+g.f'`

so, `f=x,\g=e^{x-x^2}`

`\frac{d}{dx}(x)e^{x-x^2}+\frac{d}{dx}(e^{x-x^2})x`

We know...
`d/dx(x)=1`

and,
`\frac{d}{dx}(e^{x-x^2})`
Applying chain rule, `\frac{df(u\)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}`

let `x-x^2=u`

`\frac{d}{du}(e^u\)\frac{d}{dx}\(x-x^2)`

We get,

`(e^u)(1-2x)`

Substituting back `x-x^2=u`,

=`e^{x-x^2}(1-2x)`

= `1e^{x-x^2}+e^{x-x^2}(1-2x)x`

Simplifying it,

`e^{x-x^2}(-2x^2+x+1)`

Finally when `x=-1`
`e^{(-1)-(-1)^2}[-2(-1)^2+(-1)+1]` = `e^-2 . (-2)`= `-2/e^2`